Answer :
Given:
The puma jumps to the height of: h = 3.7 m
The angle made by puma with the ground is: θ = 45°
To find:
The speed of the puma when it leaves the ground.
Explanation:
We consider only the vertical motion of the puma.
The puma jumps to the height of 3.7 m and its speed at this height will be zero.
Thus, v = 0 m/s.
The initial speed "u" can be calculated by using the following kinematical equation.
[tex]v^2=u^2+2ah[/tex]Here, a = - g. The negative sign indicates that the puma jumps against the acceleration due to gravity.
Taking g = 9.8 m/s² and substituting the values in the above equation, we get:
[tex]\begin{gathered} (0\text{ m/s\rparen}^2=u^2+2\times(-9.8\text{ m/s}^2)\times3.7\text{ m} \\ \\ 0=u^2-72.52\text{ m}^2\text{/s}^2 \\ \\ u^2=72.52\text{ m}^2\text{/s}^2 \\ \\ v=\sqrt{72.52\text{ m}^2\text{/s}^2} \\ \\ u=8.51\text{ m/s} \end{gathered}[/tex]Now, the vertical component of the speed of the puma is given as:
[tex]u=u_0sin\theta[/tex]Here, u0 is the speed with which the puma leaves the ground.
Substituting the values in the above equation, we get:
[tex]\begin{gathered} 8.51\text{ m/s}=u_0sin45\degree \\ \\ u_0=\frac{8.51\text{ m/s}}{sin45\degree} \\ \\ u_0=\frac{8.51\text{ m/s}}{0.707} \\ \\ u_0=12.03\text{ m/s} \\ \\ u_0\approx12\text{ m/s} \end{gathered}[/tex]Final answer:
The speed of the puma with which it leaves the ground is 12 m/s.